Question

In the figure, arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm, to intersect the sides BC, CA and AB at their respective mid - point D, E and F. Find the area of the shaded region $(use\pi =3.14)$

Answer 1

From the given fig, area of the shaded part is equal to the sim of areas of these sectors at points $A,B$ and $C$.

As $\Delta ABC$ is equilateral triangles of sides $10cm$ and radius of the sector is half of the side. All the three sectors are identical.

$\theta ={60}^0$

Refer image,

Radius of each sector (r) $=\frac{10}{2}=5cm$

$\therefore$ Are of shaded part =3 (Area of sector)

$=\frac{3\times \pi r^2\theta }{{360}^0}=\frac{3\times 3.14\times 5\times 5\times {60}^0}{{360}^0}$

$=1.57\times 25=39.25c{m}^2$